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Unformatted text preview: nd b such that
must be with the at R,
consistent idea as solenoid. [Note: N/271"s = n applies (m. 0)0
= 2mcos()f+msin()O,
and Eq. 5.87 <=>
Eq. 5.86. Qed
3(m.f)f  m = 3m cos() f  mcos()f+msin()O s is essentially fLoI 2
(a)m=Ia=lhrR2i.1 in circumference, so that
if the toroid is large
for s ~ R;
(
constant over the crosssection.]
47rR2 S  R2) Z,
2In(R/b)= 1 (b/R)2. I'll use a = b = R. Then A =
(b) B 5.18
Problem ~ I~; I :~2 (2 cos0 f + sin 0 8) . .
fLoI
for s
{ 1.6.2, In(s/ R) Z,
 21!' J . da is independent ;::: R. }
It doesn't matter. According to Theorem 2, in Sect.
of surface, for any given
boundary line, provided0 that J = z, f = i (for z > 0), so B ~for steadyi currents< (Eq. = 11", = i, so the field
Problem 5.26 z axis, = 0, r is divergenceless, which it is,
(c) On the
(for z 0, 0 5.31).f
Problem 5.19
is the = Kx ~ B = xfLoK Y (plus for The 0, minus for z (Eq. 5.38) reduces (for zz » R) to B ~ /LoIR2/2IzI3,
K same, with Izl3 in2place of Z3). z < exact answer &...
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This note was uploaded on 04/23/2013 for the course PHSX 531 taught by Professor Chiu during the Fall '12 term at Kansas.
 Fall '12
 chiu

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